S.chand publication New Learning Composite mathematics solution of class 8 Chapter 12 Mensuration Exercise 12D

 Exercise 12D

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Q1 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 1

Find the curved surface area and whole surface area for the cylinders shown below. Measures are in centimetres.

(a)

h=5 ,r=7, $\pi =\frac{22}{7}$

∴C.S.A=2πrh

$=2\times \frac{22}{7}\times 7\times 5$

=220cm²

T.S.A=2πrh+(πr²×2)

$=220+\frac{22}{7}\times 7\times 7\times 2$

=220+(154×2)

=220+308=528cm²

(b)

h=8 ,r=5 ,π=3.14

∴C.S.A=2πrh

=2×3.14×5×8

=251.20cm²

T.S.A=2πrh+(πr²×2)

=251.20+{(3.14×5²)×2}

=251.20+157

=408.20cm²

(c)

h=10 ,r=2, π=3.14

C.S.A=2πrh

=2×3.14×2×10

=125.60cm²

T.S.A=2πrh+2πr²

=125.6+(2×3.14×2²)

=125.6+(2×3.14×4)

=125.6+25.12=150.72cm²

(d)

h=3 ,r=7, $\pi=\frac{22}{7}$

C.S.A=2πrh

$=2\times \frac{22}{7}\times 7\times 3$

=132cm²

T.S.A=2πrh+2πr²

$=132+\left(2\times \frac{22}{7}\times 7\times 7\right)$

=132+308=440cm²

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Q2 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 2

The table given below contains information about a few cylinders. Complete it.

	Radius	Height	Curved Surface Area	Area of the base	Whole Surface
(a)	7 cm	12 cm
(b)	2.8 m	5 m
(c)	35 mm	50 mm

Sol :

(a)

r=7cm , h=12cm

T.S.A=2πrh+2πr²

$=\left(2\times \frac{22}{7}\times 7\times 12\right)+\left(2\times \frac{22}{7}\times 7 \times 7\right)$

=528+308=836cm²

Area of the base=πr²

$=\frac{22}{7}\times 7\times 7$

=154cm²

C.S.A=2πrh

$=2\times \frac{22}{7} \times 7\times 12$

=528cm²

(b)

r=2.8m ,h=5m

C.S.A=2πrh

=2×2.8×5×3.14

=87.92m²

T.S.A=2πrh+2πr²

=87.92+2×3.14×2.8×2.8

=87.92+49.22

=137.15m²

Area of the base=πr²

=3.14×2.8×2.8

=24.61m²

(c)

r=35mm ,h=50mm

C.S.A=2πrh

$=2\times \frac{22}{7}\times 35 \times 50$

=11000m²

Area of the base=πr²

$=\frac{22}{7}\times 35\times 35$

=3850m²

T.S.A=2πrh+2πr²

=11000+(2×3850)

=18700m²

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Q3 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 3

Find the missing dimension for a cylinder.

(a) Radius: Curved surface area $=220 \mathrm{~cm}^{2}$. Height =10 cm

Sol :

(a) 

C.S.A=220cm² ,h=10cm , r=?

⇒C.S.A=2πrh

$r=\frac{\text{C.S.A}}{2 \pi h}=\frac{220}{2\times \frac{22}{7}\times 10}$

$r=\frac{220\times 7}{2\times 22\times 10}=\frac{7}{2}=3.5$cm

(b) Height: Curved surface area = $3168 \mathrm{~cm}^{2}$, Radius = 21 cm

Sol :

(b)

C.S.A=2πrh

$h=\frac{\text{C.S.A}}{2\pi r}=\frac{3168}{2\times \frac{22}{7}}\times 21$

$=\frac{3168}{2\times 22\times 3}$

h=24cm

(c) Height: Radius=5 cm. Total surface area $=120 \pi\mathrm{cm}^{2}$

Sol :

(c)

T.S.A=120π , r=5 , h=?

T.S.A=2πrh+2πr²

120π=2π(rh+r²)

120π=2π(5h+5²)

60=5h+25

5h=60-25=35

h=7cm

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Q4 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 4

The circumference of a right circular cylinder is 440 cm and its height is 5 cm. Find the lateral surface area of the cylinder.

Sol :

c=440cm , h=5cm

ATQ ,

2πr=440

$r=\frac{440}{2\times \pi}$

$r=\frac{440}{2\times \frac{22}{7}}$

$=\frac{440\times 7}{2\times 22}$

=70cm

∴L.S.A=2πrh

$=2\times \frac{22}{7}\times 70\times 5$

=2200cm²

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Q5 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 5

A cylindrical soup can is 15.4 cm in diameter and 20 cm in length. What is the area of the label that covers the side of the can?

Sol :

d=15.4cm

∴$r=\frac{15.4}{2}$cm ,h=20cm

⇒Lateral Surface Area=2πrh

$=2\times \frac{22}{7}\times \frac{15.4}{2}\times 20$

$=2\times \frac{22}{7}\times \frac{154}{2\times 10}\times 20$

=968cm²

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Q6 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 6

In a building, there are 30 cylindrical pillars. The radius of each pillar is 42 m and height is 7 m. Find the total cost of painting the curved surface area of all pillars at the rate of ₹ 10 per m²?

Sol :

r=42m , h=7m

⇒C.S.A of a pillar=2πrh

$=2\times \frac{22}{7}\times 42\times 7$

=1848m²

C.S.A of 30 pillar=1848×30

=55440m²

∴Paint cost for 30 pillar's=55440×10

=554400

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Q7 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 7

A cylinder has a diameter 20 cm and height 7 cm. Explain whether tripling the height would have the same effect on the surface area as tripling the radius or not.

Sol :

d=20cm , h=7cm ,$r=\frac{20}{2}=10$cm

For 1st case:-

h₁=3×7=21cm

∴T.S.A for 1st case=2πr(r+h₁)

=2×3.14×10×(10+21)

=1946.8 cm²

For 2nd case:-

r₁=10×3=30cm

∴T.S.A to 2nd case=2πr₁(r₁+h)h

=2×3.14×30×(30+7)

=6970.8 cm²

∴Both case surfaces are will be different.

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Q8 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 8

The diameter of a garden roller is 1.8 m and it is 2.8 m long. How much area will it cover is 25 revolutions in levelling a field?

Sol :

d=1.8m , h=2.8m , $r=\frac{1.8}{2}=0.9$m

⇒C.S.A=2πrh

$=2\times \frac{22}{7}\times 0.9\times 2.8$

$=2\times \frac{22}{7}\times \frac{9}{10}\times \frac{28}{10}$

$=\frac{2\times 22\times 9\times 4}{100}=\frac{396}{25}$m²

∴1 revolution need$=\frac{396}{25}$m² area 

∴25 revolution need$=\frac{396}{25}\times 25$=396m²  

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Q9 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 9

The diameter of a road roller is 90 cm and is 2 m 10 cm in length. If it takes 400 revolutions to level a playground , find the cost of levelling the ground at 2.25 per square meter.

Sol :

d=90cm ,$r=\frac{90}{2}$=45cm

Length=2m10cm=(2×100+10)=210cm

⇒1 revolution=C.S.A of roller=2πrh

$=2\times \frac{22}{7}\times 45 \times 210$

=2×22×45×30 cm²

=59400 cm² or 5.94 m² 

If takes 400 revolution to level 5.94 m² area

It takes 400 revolutions , so total area of ground = 5.94×400=2376 m²

∴Cost of levelling at 2.25 for 1m²

∴Cost of levelling 

(2376×2.25)₹=(2376×1) m²

5346 ₹=2376 m²

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Q10 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 10

The inner diameter of a 15 m deep circular well is 4.2 m. Find the cost of plastering its inner covered surface at the rate of 40 per m²

Sol :

Inner diagram=4.2m or radius$=\frac{4.2}{2}$=2.1m

Height(h)=15m

⇒Inner C.S.A=2πrh

$=2\times \frac{22}{7}\times 2.1\times 15$

$=2\times \frac{22}{7}\times \frac{21}{10}\times 15$

=198m²

Cost of plastering of 1m² area=40

∴Cost of plastering of 198 m² area=198×40=7920

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Q11 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 11

The inner circumference of a hollow cylindrical pipe is 4.5 m and outer circumference is 4.6 m. Find the cost of painting both its inner and outer sides at the rate of 20 per m², if it is 8 m tall

Sol :

Inner Circumference=4.5m

Outer Circumference=4.6m

Height(h)=8m

⇒Inner Circumference=2πr=4.5

∴$r=\frac{4.5}{2\pi}$

$=\frac{4.5}{2\times 3.14}=\frac{4.5}{6.28}$

=0.715m

Outer circumference=2πR=4.6

$R=\frac{4.6}{2\pi}=\frac{4.6}{2\times 3.14}$

=0.731m

Inner C.S.A=2πrh

$=2\times \frac{22}{7}\times 0.731\times 8$

=35.95m²

Outer C.S.A=2πRh

$=2\times \frac{22}{7}\times 0.731\times 8$

=36.75m²

∴Total C.S.A=35.95+36.75=72.7m²

∵Cost of painting 1m²=20

∴Cost of painting hollow cylinder=72.7×20=1454

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Q12 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 12

The external diameter of a 20 cm long and 1 cm thick hollow iron pipe is 25 cm. Determine the whole surface of the iron pipe.

Sol :

D=25cm ,t=1cm , $R=\frac{25}{2}=12.5$cm

r=12.5-1=11.5cm

h=20cm

C.S.A of Iron pipe=2πh(R+r)

$=2\times \frac{22}{7}\times 20(12.5+11.5)$

$=2\times \frac{22}{7}\times 20 \times 24$

$=\frac{44\times 480}{7}$ cm²

∴Area of one circular end=2(πR²-πr²)

$=2\times \frac{22}{7} \left\{(12.5+11.5)(12.5-11.5)\right\}$

$=\frac{44}{7}\times 24$ cm²

∴T.S.A of Iron pipe$=\left(\frac{44\times 480}{7}\right)+\left(\frac{44}{7}\times 24\right)$

$=\frac{44}{7}\left(480+24\right)$

$=\frac{44}{7}\times 504$

=44×72=3168 cm²